$\newcommand{\g}{\mathfrak{g}}$ $\newcommand{\h}{\mathfrak{h}}$ Let $\mathfrak g$ be a a lie algebra with a given cartan matrix $A$, invariant billinear form $(|)$ and root space decomposition $ \mathfrak g=\h \oplus_{\alpha > 0 } (g_\alpha \oplus g_{-\alpha})$, basis $e_\alpha^i$ of the $g_\alpha$ for $\alpha>0$ and dual basis $e_{-\alpha}$ with respect to $(,)$. Let there also be given a basis $u_i$ of the cartan $\h$, and a dual basis $u^i$ with respect to $(|)$. Let $\langle , \rangle$ denote the evaluation pairing of $\h^*$ on $\h$. Multiplication denotes as usual multiplication in the universal enveloping algebra of $\g$
Question: Kac, on page 23 of Infinite dimensional lie algebras writes a step I don't understand:
$\sum_i \langle \alpha, u^i \rangle xu_i+ \sum_i u^i \langle \alpha, u_i \rangle x=\sum_i \langle \alpha, u^i \rangle \langle \alpha, u_i \rangle x + x\left( \sum_i u^i \langle \alpha, u_i \rangle + u_i \langle \alpha, u^i \rangle \right)$.
I am skeptical of this equality because $\sum_i \langle \alpha, u^i \rangle \langle \alpha, u_i \rangle x$ from the RHS is of degree one in $U(\g)$, whereas the left hand side is purely of degree 2.
Why is this step true?
Silly me: Just use the fact that $x \otimes y=y \otimes x + [x,y]$.